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%I A292465 #28 Sep 08 2022 08:46:19
%S A292465 0,1,4,18,60,200,624,1911,5712,16830,48950,140976,402624,1141933,
%T A292465 3219580,9031050,25219824,70153016,194466672,537404835,1480993800,
%U A292465 4071156726,11165970794,30561658848,83490220800,227687745625,619938027124,1685442626946,4575973716132
%N A292465 a(n) = n*F(n)*F(n+1), where F = A000045.
%C A292465 Inequality proposed by Bătineţu-Giurgiu and Stanciu (see References): Let {x(n)}_{n>=1} be a sequence of real numbers. Prove that 2*(Sum_{k=1..n} F(k)*sin(x(k)))*(Sum_{k=1..n} F(k)*cos(x(k))) <= n*F(n)*F(n+1).
%H A292465 Vincenzo Librandi, <a href="/A292465/b292465.txt">Table of n, a(n) for n = 0..1000</a>
%H A292465 D. M. Bătineţu-Giurgiu and N. Stanciu, <a href="http://www.fq.math.ca/Problems/ElemProbSolnNov2015.pdf">Problem B-1179</a>, The Fibonacci Quarterly, Volume 53, Number 4 (November 2015), p. 366.
%H A292465 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,-6,20,-6,-9,6,-1).
%F A292465 G.f.: x*(1 - 2*x + 3*x^2 - 6*x^3 + 6*x^4 - 2*x^5)/((1 - x)^2*(1 + x)^2*(1 - 3*x + x^2)^2).
%p A292465 with(combinat,fibonacci): A292465:=seq(n*fibonacci(n)*fibonacci(n+1), n=0..10^2); # _Muniru A Asiru_, Sep 26 2017
%t A292465 Table[n Fibonacci[n] Fibonacci[n+1], {n, 0, 30}]
%o A292465 (Magma) [n*Fibonacci(n)*Fibonacci(n+1): n in [0..35]];
%o A292465 (PARI) a(n) = n*fibonacci(n)*fibonacci(n+1); \\ _Altug Alkan_, Sep 17 2017
%o A292465 (GAP)
%o A292465 A292465:=List([0..10^3],n->n*Fibonacci(n)*Fibonacci(n+1)); # _Muniru A Asiru_, Sep 26 2017
%Y A292465 Cf. A000045, A001654.
%K A292465 nonn,easy
%O A292465 0,3
%A A292465 _Vincenzo Librandi_, Sep 17 2017

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